Answer
The only combination of $\epsilon_0$ and $\mu_0$ with the dimensions of speed is $(\epsilon_0~\mu_0)^{-\frac{1}{2}}$
Work Step by Step
$\epsilon_0 = 8.85\times 10^{-12}~m^{-3}~kg^{-1}~s^4~A^2$
$\mu_0 = 1.26\times 10^{-6}~m~kg~s^{-2}~A^{-2}$
Speed $v$ has units of m/s.
Suppose $v = C~\epsilon_0^a~\mu_0^b$, where $C$ is some dimensionless constant.
Consider the units of kilograms:
$(kg^{-1})^a~(kg)^b = kg^0$
$b-a = 0$
$a = b$
Consider the units of meters:
$(m^{-3})^a~(m)^b = m^1$
$b - 3a = 1$
$a - 3a = 1$
$a = -\frac{1}{2}$
Then $b = -\frac{1}{2}$, since $a = b$
$v = C~\epsilon_0^a~\mu_0^b$
$v = C~\epsilon_0^{-\frac{1}{2}}~\mu_0^{-\frac{1}{2}}$
$v = C~(\epsilon_0~\mu_0)^{-\frac{1}{2}}$
The only combination of $\epsilon_0$ and $\mu_0$ with the dimensions of speed is $(\epsilon_0~\mu_0)^{-\frac{1}{2}}$
